In Fig. 6.7, ∠D = ∠E and AD/DB = AE/EC. Prove that BAC is an isosceles triangle
Solution:
Given, ∠D = ∠E
Also, AD/AB = AE/EC
We have to prove that BAC is an isosceles triangle.
Basic Proportionality Theorem states(BPT) that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given, AD/AB = AE/EC
By BPT, DE is parallel to BC i.e., DE||BC
Also, the corresponding angles are equal.
So, ∠D = ∠B and ∠C = ∠E
Given, ∠D = ∠E
So, ∠B = ∠C
It is clear that the sides AB and AC are equal.
Isosceles triangle theorem states that in an isosceles triangle, if two sides of a triangle are congruent, then angles opposite to those sides are congruent.
Therefore, BAC is an isosceles triangle.
✦ Try This: D and E are the points that lie on the sides PQ and PR of the triangle PQR. ∠D = ∠E and DE||QR Prove that PQR is an isosceles triangle
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.3 Sample Problem 3
In Fig. 6.7, ∠D = ∠E and AD/DB = AE/EC. Prove that BAC is an isosceles triangle
Summary:
In Fig. 6.7, ∠D = ∠E and AD/AB = AE/EC. It is proved that BAC is an isosceles triangle
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