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Apr 20, 2019

Problems on Size Transformation | Area of the Terrace in the Model

Here we will solve different types of problems on size transformation. 1. A map of a rectangular park is drawn to a scale of 1 : 5000. (i) Find the actual length of the park if the length of the same in the map is 25 cm. (ii) If the actual width of the park is 1 km, find its

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Apr 20, 2019

Reduction Transformation | Centre of Reduction | Reduction Factor

We will discuss here about the similarity on Reduction transformation. In the figure given below ∆X’Y’Z’ is a reduced image of ∆XYZ. The two triangles are similar. Here also the triangles are equiangular and \(\frac{X’Y’}{XY}\) = \(\frac{Y’Z’}{YZ}\) = \(\frac{Z’X’}{ZX}\) = k

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Apr 20, 2019

Enlargement Transformation | Centre of Enlargement |Enlargement Factor

We will discuss here about the similarity on enlargement transformation. Cut out some geometrical figures like triangles, quadrilaterals, etc., from a piece of cardboard. Hold these figures, one-by-one, between a point source of light and a wall.

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Apr 18, 2019

AA Criterion of Similarly on Quadrilateral | Alternate Angles

Here we will prove that in the quadrilateral ABCD, AB ∥ CD. Prove that OA × OD = OB × OC. Solution: Proof: 1. In ∆ OAB and ∆OCD, (i) ∠AOB = ∠COD (ii) ∠OBA = ∠ODC. 2. ∆ OAB ∼ ∆OCD. 3. Therefore, OA/OC = OB/OD ⟹ OA × OD = OB × OC. (Proved)

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Apr 13, 2019

Greater segment of the Hypotenuse = the Smaller Side of the Triangle

Here we will prove that if a perpendicular is drawn from the right-angled vertex of right-angled triangle to the hypotenuse and if the sides of the right-angled triangle are in continued proportion, the greater segment of the hypotenuse is equal to the smaller side of the

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Apr 06, 2019

Application of Basic Proportionality Theorem | Internal Bisector

Here we will prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. Given: XP is the internal bisector of ∠YXZ, intersecting YZ at P.

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Apr 04, 2019

Converse of Basic Proportionality Theorem | Proof with Diagram

Here we will prove converse of basic proportionality theorem. The line dividing two sides of a triangle proportionally is parallel to the third side. Given: In ∆XYZ, P and Q are points on XY and XZ respectively, such that XP/PY = XQ/QZ. To prove: PQ ∥ YZ Proof: Statement

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Mar 28, 2019

Basic Proportionality Theorem | AA Criterion of Similarity | Diagram

Here we will learn how to prove the basic proportionality theorem with diagram. A line drawn parallel to one side of a triangle divides the other two sides proportionally. Given In ∆XYZ, P and Q are points on XY and XZ respectively, such that PQ ∥ YZ. To prove XP/PY = XQ/QZ.

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Mar 28, 2019

Criteria of Similarity between Triangles | SAS Criterion of Similarity

We will discuss here about the different criteria of similarity between triangles with the figures. 1. SAS criterion of similarity: If two triangles have an angle of one equal to an angle of the other and the sides including them are proportional, the triangles are similar.

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Mar 28, 2019

Similar Triangles | Congruency and Similarity of Triangles | Diagram

We will discuss here about the similar triangles. If two triangles are similar then their corresponding angles are equal and corresponding sides are proportional. Here, the two triangles XYZ and PQR are similar. So, ∠X = ∠P, ∠Y = ∠Q, ∠Z = ∠R and XY/PQ=YZ/QR = XZ/PR. ∆XYZ is

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Mar 25, 2019

AA Criterion of Similarity | Similarity on Right-angled Triangle

Here we will prove that in a right-angled triangle, if a perpendicular is drawn from the right-angled vertex to the hypotenuse, the triangles on each side of it are similar to the whole triangle and to one another. Given: Let XYZ be a right angle in which ∠YXZ = 90° and

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Mar 19, 2019

Properties of size Transformation |Enlargement|Reduction|Scale Factor

We will discuss here about the different properties of size transformation. 1. The shape of the image is the same as that of the object. 2. If the scale factor of the transformation is k then each side of the image is k times the corresponding side of the object.

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Mar 08, 2019

Proof By the Equal Intercepts Theorem | Line Joining the Midpoints

Here we will prove that in the given ∆XYZ, M and N are the midpoints of XY and XZ respectively. T is any point on the base YZ. Prove that MN bisects XT. Solution: Given: In ∆XYZ, XM = MY and XN = NZ. MN cuts XT at U. To prove: XU = UT. Construction: Through X, draw PQ ∥ YZ.

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Mar 08, 2019

Midpoint Theorem by using the Equal Intercepts Theorem |Proof |Diagram

Here we will prove that converse of the Midpoint Theorem by using the Equal Intercepts Theorem. Solution: Given: P is the midpoint of XY in ∆XYZ. PQ ∥ YZ. To prove: XQ = QZ. Construction: Through X, draw MN ∥ YZ. Proof: Statement 1. PQ ∥ YZ. 2. MN ∥ PQ ∥ YZ.

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Mar 05, 2019

Problems on Equal Intercepts Theorem | Midpoint Theorem | Diagram

Here we will solve different types of problems on Equal Intercepts Theorem. 1. In the given figure, MN ∥ KL ∥ GH and PQ = QR. If ST = 2.2 cm, find SU. Solution: The transversal PR makes equal intercepts, PQ and QR, on the three parallel lines MN, KL and GH. Therefore, by the

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Feb 26, 2019

Equal Intercepts Theorem | Transversal makes Equal Intercepts

Intercept In the figure given above, XY is a transversal cutting the line L1 and L2 at P and Q respectively. The line segment PQ is called the intercept made on the transversal XY by the lines L1 and L2. If a transversal makes equal intercepts on three or more parallel lines

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Feb 25, 2019

Collinear Points Proved by Midpoint Theorem | Collinearity | Diagram

In ∆XYZ, the medians ZM and YN are produced to P and Q respectively such that ZM = MP and YN = NQ. Prove that the points P, X and Q are collinear, and X is the midpoint of PQ. Solution: Given: In ∆XYZ, the points M and N are the midpoints of XY and XZ respectively.

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Feb 22, 2019

Midpoint Theorem on Right-angled Triangle | Proof | Statement | Reason

Here we will prove that in a right-angled triangle the median drawn to the hypotenuse is half the hypotenuse in length. Solution: In ∆PQR, ∠Q = 90°. QD is the median drawn to hypotenuse PR

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Feb 21, 2019

Midsegment Theorem on Trapezium | Nonparallel Sides of a Trapezium

Here we will prove that the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them. Solution: Given: PQRS is a trapezium in which PQ ∥ RS. U and V are the

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Feb 20, 2019

Midpoint Theorem on Trapezium | Converse of the Midpoint Theorem

PQRS is a trapezium in which PQ ∥ RS. T is the midpoint of QR. TU is drawn parallel to PQ which meets PS at U. Prove that 2TU = PQ + RS. Given: PQRS is a trapezium in which PQ ∥ RS. T is the midpoint of QR. TU ∥ PQ and TU meets PS at U. To prove: 2TU = PQ + RS. Construction

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Feb 13, 2019

Straight Line Drawn from the Vertex of a Triangle to the Base |Diagram

Here we will prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other two sides of the triangle. Solution: Given: Q and R are the midpoints of the sides XY and XZ respectively of

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Feb 12, 2019

Four Triangles which are Congruent to One Another | Prove with Diagram

Here we will show that the three line segments which join the middle points of the sides of a triangle, divide it into four triangles which are congruent to one another. Solution: Given: In ∆PQR, L, M and N are the midpoints of QR, RP and PQ respectively. To prove ∆PMN ≅ LNM

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Feb 11, 2019

Midpoint Theorem Problem | Midpoint Theorem | Converse of Midpoint

Here we will learn how to solve different types of midpoint theorem problem. In the adjoining figure, find (i) ∠QPR, (ii) PQ if ST = 2.1 cm. Solution: In ∆PQR, S and T are the midpoints of PR and QR respectively. Therefore, ST = \(\frac{1}{2}\)PQ and ST ∥ PQ.

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Feb 06, 2019

Converse of Midpoint Theorem | Proof of Converse of Midpoint Theorem

The straight line drawn through the midpoint of one side of a triangle parallel to another bisects the third side. Given: In ∆PQR, S is the midpoint of PQ, and ST is drawn parallel to QR. To prove: ST bisects PR, i.e., PT = TR. Construction: Join SU where U is the midpoint

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Feb 04, 2019

Midpoint Theorem |AAS & SAS Criterion of Congruency Prove with Diagram

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. Given: A triangle PQR in which S and T are the midpoint of PQ and PR respectively. To prove: ST ∥ QR and ST = 1/2QR Construction: Draw RU ∥ QP such that

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Jan 31, 2019

Sum of Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals

Here we will prove that in any quadrilateral the sum of the four sides exceeds the sum of the diagonals. Solution: Given: ABCD is a quadrilateral; AC and BD are its diagonals. To prove: (AB + BC + CD + DA) > (AC + BD). Proof: Statement 1. In ∆ADB, (DA + AB) > BD.

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Jan 30, 2019

Sum Of Any Two Sides Is Greater Than Twice The Median | Proof|Diagram

Here we will prove that in a triangle the sum of any two sides is greater than twice the median which bisects the remaining side. Solution: Given: In ∆XYZ, XP is the median that bisects YZ at P. To prove: (XY + XZ) > 2XP. Construction: Produce XP to Q such that XP = PQ.

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Jan 28, 2019

Problem on Inequalities in Triangle | Solution with Diagram

Here we will solve the problem on inequalities in triangle. Let XYZ be a triangle in which XM bisects ∠YXZ. Prove that XY is greater than YM. As XM bisects ∠YXZ, we have ∠YXZ = ∠MXZ ............ (i) Also, in ∆XMZ, ∠XMY > ∠MXZ, as an exterior angle of a triangle is always

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Jan 25, 2019

Comparison of Sides and Angles in a Triangle | Geometrical Property

Here we will solve different types of problems on comparison of sides and angles in a triangle. 1. In ∆XYZ, ∠XYZ = 35° and ∠YXZ = 63°. Arrange the sides of the triangle in the descending order of their lengths. ∠XZY = 180° - (∠XYZ + ∠YXZ) = 180° - (35° + 63°) = 180° - 98°

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Jan 24, 2019

Perpendicular is the Shortest Theorem | Inequalities in Triangle

Here we will prove that of all the straight lines that can be drawn to a straight line from a given point outside it, the perpendicular is the shortest. Given: XY is a straight line and O is a point outside it. OP is perpendicular to XY and OZ is an oblique. To Prove: OP

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Jan 23, 2019

The Sum of any Two Sides of a Triangle is Greater than the Third Side

Here we will prove that the sum of any two sides of a triangle is greater than the third side. Given: XYZ is a triangle. To Prove: (XY + XZ) > YZ, (YZ + XZ) > XY and (XY + YZ) > XZ Construction: Produce YX to P such that XP = XZ. Join P and Z. Statement 1. ∠XZP = ∠XPZ.

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Jan 21, 2019

Greater Side has the Greater Angle Opposite to It | Triangle Inequalit

Here we will prove that if two sides of a triangle are unequal, the greater side has the greater angle opposite to it. Given: In ∆XYZ, XZ > XY To prove: ∠XYZ > ∠XZY. Construction: From XZ, cut off XP such that XP equals XY. Join Y and P. Proof: Statement 1. In ∆XYP, ∠XYP =

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Jan 21, 2019

Greater Angle has the Greater Side Opposite to It | Prove with Diagram

Here we will prove that if two angles of a triangle are unequal, the greater angle has the greater side opposite to it. Given: In ∆XYZ, ∠XYZ > ∠XZY To Prove: XZ > XY Proof: Statement 1. Let us assume that XZ is not greater than XY. Then XZ must be either equal to or less

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Jan 17, 2019

Theorem on Isosceles Triangle | Proof Involving Isosceles Triangles

Here we will prove that the equal sides YX and ZX of an isosceles triangle XYZ are produced beyond the vertex X to the points P and Q such that XP is equal to XQ. QY and PZ are joined. Show that QY is equal to PZ. Solution: In ∆XYZ, XY = XZ. YX and XZ are produced to P and

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Jan 16, 2019

Points on the Base of an Isosceles Triangle | Prove with Diagram

Here we will prove that if two given points on the base of an isosceles triangle are equidistant from the extremities of the base, show that they are also equidistance from the vertex. Solution: Given: In the isosceles ∆XYZ, XY = XZ, M and N points on the base YZ such that

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Jan 15, 2019

Lines Joining the Extremities of the Base of an Isosceles Triangle

Here we will show that the straight lines joining the extremities of the base of an isosceles triangle to the midpoints of the opposite sides are equal. Solution: Given: In ∆XYZ, XY = XZ, M and N are the midpoints of XY and XZ respectively.

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Jan 10, 2019

Problem on Two Isosceles Triangles on the Same Base | Proof | Diagram

Here we will prove that ∆PQR and ∆SQR are two isosceles triangles drawn on the same base QR and on the same side of it. If P and S be joined, prove that each of the angles ∠QPR and ∠QSR will be divided by the line PS into two equal parts.

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Jan 09, 2019

Problems on Properties of Isosceles Triangles | Find x° and y°

Here we will solve some numerical problems on the properties of isosceles triangles Find x° from the given figures. In ∆XYZ, XY = XZ. Therefore, ∠XYZ = ∠XZY = x°. Now, ∠YXZ + ∠XYZ + XZY = 180° ⟹ 84° + x° + x° = 180° ⟹ 2x° = 180° - 84° ⟹ 2x° = 96°

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Jan 07, 2019

Three Angles of an Equilateral Triangle are Equal | Axis of Symmetry

Here we will prove that if the three angles of a triangle are equal, it is an equilateral triangle. Given: In ∆XYZ, ∠YXZ = ∠XYZ = ∠XZY. To prove: XY = YZ = ZX. Proof: Statement 1. XY = ZX. 2. XY = YZ. 3. XY = YZ = ZX. (Proved) Reason 1. Sides opposite to equal angles ∠XZY

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Jan 05, 2019

Sides Opposite to the Equal Angles of a Triangle are Equal | Diagram

Here we will prove that the sides opposite to the equal angles of a triangle are equal. Given: In ∆ABC, ∠XYZ = ∠XZY. To prove: XY = XZ. Construction: Draw the bisector XM of ∠YXZ so that it meets YZ at M. Proof: Statement 1. In ∆XYM and ∆XZM, (i) ∠XYM = XZM (ii) ∠YXM = ∠ZXM

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Jan 03, 2019

The Three Angles of an Equilateral Triangle are Equal | With Diagram

Here we will prove that the three angles of an equilateral triangle are equal. Given: PQR is an equilateral triangle. To prove: ∠QPR = ∠PQR = ∠ PRQ. Proof: Statement 1. ∠QPR = ∠PQR 2. ∠PQR = ∠ PRQ. 3. ∠QPR = ∠PQR = ∠ PRQ. (Proved). Reason 1. Angles opposite to equal sides QR

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Jan 02, 2019

Equal Sides of an Isosceles Triangle are Produced, the Exterior Angles

Here we will prove if the equal sides of an isosceles triangle are produced, the exterior angles are equal. Given: In the isosceles triangle PQR, the equal sides PQ and PR are produced to S and T respectively. To prove: ∠RQS = ∠QRT. Proof: Statement 1. ∠PQR = ∠PRQ 2. ∠RQS

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Jan 01, 2019

Angles Opposite to Equal Sides of an Isosceles Triangle are Equal

Here we will prove that in an isosceles triangle, the angles opposite to the equal sides are equal. Solution: Given: In the isosceles ∆XYZ, XY = XZ. To prove ∠XYZ = ∠XZY. Construction: Draw a line XM such that it bisects ∠YXZ and meets the side YZ at M. Proof: Statement

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Dec 31, 2018

Application of Congruency of Triangles | Isosceles triangle Proved

Here we will prove some Application of congruency of triangles. PQRS is a rectangle and POQ an equilateral triangle. Prove that SRO is an isosceles triangle. Solution: Given: PQRS is a rectangle. POQ is an equilateral triangle to prove ∆SOR is an isosceles triangle. Proof:

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Dec 28, 2018

Prove that the Bisectors of the Angles of a Triangle Meet at a Point

Here we will prove that the bisectors of the angles of a triangle meet at a point. Solution: Given In ∆XYZ, XO and YO bisect ∠YXZ and ∠XYZ respectively. To prove: OZ bisects ∠XZY. Construction: Draw OA ⊥ YZ, OB ⊥ XZ and OC ⊥ XY. Proof: Statement 1. In ∆XOC and ∆XOB

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Dec 27, 2018

Point on the Bisector of an Angle | Corresponding Parts of a Triangles

Here we will prove that any point on the bisector of an angle is equidistant from the arms of that angle. Solution: Given OZ bisects ∠XOY and PM ⊥ XO and PN ⊥ OY. To prove PM = PN. Proof: Statement 1. In ∆OPM and ∆OPN, (i) ∠MOP = ∠NOP. (ii) ∠OMP = ∠ONP = 90°

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Dec 27, 2018

Prove that an Altitude of an Equilateral Triangle is also a Median

Here we will prove that an altitude of an equilateral triangle is also a median. In a ∆PQR, PQ = PR. Prove that the altitude PS is also a medina. Solution: Given in ∆PQR, PQ = PR and PS ⊥ QR.To prove PS is a median, i.e., QS = SR Proof: Statement 1. In ∆PQS and ∆PRS,

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Dec 24, 2018

Problems on Congruency of Triangles |Prove Two Triangles are Congruent

Here we will learn how to prove different types of problems on congruency of triangles. 1. PQR and XYZ are two triangles in which PQ = XY and ∠PRQ = 70, ∠PQR = 50°, ∠XYZ = 70°, and ∠YXZ = 60°. Prove that the two triangles are congruent. Solution: In a triangle, the sum of

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Dec 20, 2018

Criteria for Congruency | SAS| AAS | SSS | RHS | CPCTC

Here we will learn different criteria for congruency of triangles. I. SAS (Side-Angle-Side) Criterion: If two triangles have two sides of one equal to two sides of the other, each to each, and the angles included by those sides are equal then the triangles are congruent.

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