In the adjoining figure, points D and E are on side BC of ∆ABC, such

1.	In the adjoining figure, points D and E are on side BC of ∆ABC, such that BD = CE and AD AE. Show that ∆ABD ≅ ∆ACE.
Answer» Given: Points D and E are on side BC of ∆ABC, such that BD = CE and AD = AE. To prove: ∆ABD ≅ ∆ACE Proof: In ∆ADE, seg AD = seg AE [Given] ∴ ∠AED = ∠ADE …(i) [Isosceles triangle theorem] Now, ∠ADE + ∠ADB = 180° …(ii) [Angles in a linear pair] ∴ ∠AED + ∠AEC = 180° ….(iii) [Angles in a linear pair] ∴ ∠ADE + ∠ADB = ∠AED + ∠AEC [From (ii) and (iii)] ∴ ∠ADE + ∠ADB = ∠ADE + ∠AEC [From (i)] ∴ ∠ADB = ∠AEC ….(iv) [Eliminating ∠ADE from both sides] In ∆ABD and ∆ACE, seg BD ≅ seg CE [Given] ∠ADB = ∠AEC [From (iv)] seg AD ≅ seg AE [Given] ∴ ∆ABD ≅ ∆ACE [SAS test]

In the adjoining figure, points D and E are on side BC of ∆ABC, such that BD = CE and AD AE. Show that ∆ABD ≅ ∆ACE.

Answer»

Given: Points D and E are on side BC of ∆ABC, such that BD = CE and AD = AE.

To prove: ∆ABD ≅ ∆ACE

Proof:

In ∆ADE, seg AD = seg AE [Given]

∴ ∠AED = ∠ADE …(i) [Isosceles triangle theorem]

Now, ∠ADE + ∠ADB = 180° …(ii) [Angles in a linear pair]

∴ ∠AED + ∠AEC = 180° ….(iii) [Angles in a linear pair]

∴ ∠ADE + ∠ADB = ∠AED + ∠AEC [From (ii) and (iii)]

∴ ∠ADE + ∠ADB = ∠ADE + ∠AEC [From (i)]

∴ ∠ADB = ∠AEC ….(iv) [Eliminating ∠ADE from both sides]

In ∆ABD and ∆ACE, seg BD ≅ seg CE [Given]

∠ADB = ∠AEC [From (iv)]

seg AD ≅ seg AE [Given]

∴ ∆ABD ≅ ∆ACE [SAS test]