103 h a AABCM is the midpoint of BC P is any paint on AM and

103. h a AABC,M is the mid-point of BC, P is any paint on AM and PE, PF are perpendiculars to AB, AC respectively. If EF is parallel to BC, Find out the value of

Solution

Angle A=90 degree

Because EF is parallel to BC which is only possible if E and F are the side AB and AC respectively.

Thus intersection point of AM, CE and FC are perpendicular bisector also.

Also AM, CE and FC are angle bisector from congruency by RHS.

Hence angle A=90 degree.