CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

$ AB$ is a line segment and$ P$ is its mid-point. $ D$ and $ E$ are points on the same side of $ AB$ such that $ \angle BAD =\angle ABE$ and $ \angle EPA = \angle DPB$ (see Fig.) .

Show that (i) $ \Delta DAP\cong \Delta EBP$ (ii) $ AD = BE$

Open in App
Solution

Given

P is the midpoint of line segment AB.

BAD=ABE and EPA=DPB

To Prove

(i) DAPEBP

(ii) AD=BE

Proof

BAD=ABE and EPA=DPB

On adding DPEon both the sides of two equal angle EPA=DPB, we get

EPA+DPE=DPB+DPE

DPA=EPB

From DAP and EBP

DPA=EPB {from above }…… (i)

AP=BP { Given }……………… (ii)

BAD=ABE { given }…………..(iii)

From above three equation both the triangle satisfies ASA congruence criterion

So, ΔDAPΔEBP

(ii) AD andBEare equal as they are corresponding parts of congruent triangles(CPCT).

So that, AD=BE

Hence Proved.


flag
Suggest Corrections
thumbs-up
2
BNAT
mid-banner-image