# In fig. 6.20, l||m and line segments AB, CD and EF are concurrent at point P. Prove that AE/BF = AC/BD = CE/FD

**Solution:**

Given, l || m

The line segments AB, CD and EF are concurrent at point P.

We have to prove that AE/BF = AC/BD = CE/FD

The points A, E and C lie on l.

The points D, F and B lie on m.

In △APC and △PDB,

Vertically opposite angles are equal. i.e.,∠APC = ∠DPB

Alternate angles are equal i.e., ∠PAC = ∠PBD

AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.

By AAA criterion, △APC ⩬ △PDB

By the property of similarity,

The corresponding sides are proportional.

AP/PB = PC/PD = AC/DB -------------- (1)

In △APE and △PBF,

Vertically opposite angles are equal. i.e.,∠APE = ∠BPF

Alternate angles are equal i.e., ∠PAE = ∠PBF

AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.

By AAA criterion, △APE ⩬ △BPF

By the property of similarity,

The corresponding sides are proportional.

AP/PB = AE/FB = EP/FP ------------------------ (2)

In △CPE and △PDF,

Vertically opposite angles are equal. i.e.,∠CPE = ∠DPF

Alternate angles are equal i.e., ∠PCE = ∠PDF

AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.

By AAA criterion, △CPE ⩬ △PDF

By the property of similarity,

The corresponding sides are proportional.

So, EP/FP = PC/PD = CE/DF ----------------- (3)

Equating (1), (2) and (3), we get,

AP/PB = PC/PD = AC/DB = AP/PB = AE/FB = EP/FP = EP/FP = PC/PD = CE/DF

Cancelling out common terms,

AC/DB = AP/PB = AE/FB = CE/DF = EP/FP ------------------------- (4)

From (4), AE/BF = AC/BD = CE/FD

Therefore, it is proved that AE/BF = AC/BD = CE/FD

**✦ Try This: **In the adjoining figure, ABCD is a ||gm in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 6

**NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 13**

## In fig. 6.20, l||m and line segments AB, CD and EF are concurrent at point P. Prove that AE/BF = AC/BD = CE/FD

**Summary:**

In fig. 6.20, l||m and line segments AB, CD and EF are concurrent at point P. It is proven that AE/BF = AC/BD = CE/FD by AAA criterion which states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal

**☛ Related Questions:**

- In Fig. 6.21, PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm . . . .
- O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through . . . .
- In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is . . . .

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