Question

From RD SHARMA: pg no. 13.63 ex. 13.4 level.1 , Q.9:
In fig. 13.97, AB=AC & CP ║ BA and AP is the bisector of exterior Please answer this Question....i will mark ur answer as brainliest and give u 50 points

Answer 1

Compare triangles CED and CAB for similarity.

You've got angle ECD of CED equal to angle ACB of CAB.

Then you've got angle CED equal to angle CAB as both are corresponding angles.

Same case with angles CDE and CBA.

By AAA test or even AA test of similarity, we can prove that the triangles CED and CAB are similar.

Thus by the laws of similar triangles, we can say that CD/CB = CE/CA.

But CD/CB = 1/2 as AD is a median on side BC.

Thus CE/CA = 1/2, and hence E is mid point of side CA. Thus we can prove that BE is another median.