In Fig.7.4, D and E are points on side BC of a ∆ ABC such that BD = CE and AD = AE. Show that ∆ ABD ≅ ∆ ACE.
Solution:
Given, ABC is a triangle.
D and E are the points on the side BC of the triangle.
BD = CE
AD = AE
We have to show that the triangles ABD and ACE are congruent.
We know that the angles opposite to the equal sides are equal.
Given, AD = AE
So, ∠ADE = ∠AED -------------------- (1)
We know that the linear pair of angles is equal to 180 degrees.
∠ADB + ∠ADE = 180°
∠ADB = 180° - ∠ADE
From (1),
∠ADB = 180° - ∠AED -------------------- (2)
∠AEC + ∠AED = 180°
∠AEC = 180° - ∠AED
From (2),
∠ADB = ∠AEC
Considering triangles ABD and ACE,
BD = CE (given)
AD = AE (given)
∠ADB = ∠AEC
By SAS criterion, ΔABD ≅ΔACE
✦ Try This: In figure, if DE || BC, then find the ratio of ar (Δ ADE) and ar (DECB).
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.3 Problem 2
In Fig.7.4, D and E are points on side BC of a ∆ ABC such that BD = CE and AD = AE. Show that ∆ ABD ≅ ∆ ACE
Summary:
In Fig.7.4, D and E are points on side BC of a ∆ ABC such that BD = CE and AD = AE. It is shown that ∆ ABD ≅ ∆ ACE by SAS criterion
☛ Related Questions: