Question

answer please because i dont know

Answer 1

Answer:

1. (i)

Solve:- In ∆ABD and ∆ACD,we get

AB=AC. (∆ABC is isosceles).

DB=DC. (∆DBC is isosceles).

and AD=AD. (common).

therefore, ∆ABD congruent to ∆ACD. (by SSS congruency).

so, angle BAD = angle CAD.(by cpct).

(ii) In ∆ABP and ∆ACP, we get

AB=AC. (sides of an isosceles∆).

angle BAP= angle CAP. ( angle BAD=angle CAD).

and AP=AP. (common).

therefore, ∆ABP congruent to ∆ACP. ( by SAS criteria).

2. In ∆ABD and ∆ACD,we get

AB=AC. (given).

AD=AD. (common).

angle ADB=angle ADC. (each equal to 90°).

therefore, ∆ABD congruent to ∆ACD. (by RHS criteria).

so, BD=DC. (by cpct).

By this we can say that AD bisects BC.

and, angle BAD=angle CAD. (by cpct).

By this we can say that AD bisects angle A.

4. In ∆BFC and ∆CEB, we get

CF=BE. (given).

angle CFB=angle BEC. (each equal to 90°).

BC=BC. (common).

therefore, ∆BFC congruent to ∆CEB. (by RHS criteria).

so, angle FBC = angle ECB. (by cpct).

therefore, AB = AC. (sides opposite to equal angles are equal).

Hence, by this we can say that ∆ABC is an isosceles triangle.

Hope this helps you.

Thankyou.