1. the perpendicular bisector of side ab of ∆abc intersects side ab at point d and bc at point e. if m∠cab = 82° and m∠c = 68°, find m∠cae.(not 14)2. the perpendicular bisectors of sides ac and bc of δabc intersect side ab at points p and q respectively, and intersect each other in the exterior (outside) of δabc. find the measure of ∠acb if m∠cpq = 78° and m∠cqp = 62°.

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1. The sum of angles of ΔABC is 180°, so ∠B is ...

... ∠B = 180° - ∠A - ∠C = 180° -82° -68° = 30°

ΔABE is isosceles, so ∠EAB = ∠B = 30°.

The sum of angles CAE and EAB will be the measure of ∠A, so we have ...

... ∠CAE + ∠EAB = ∠A

... ∠CAE + 30° = 82° . . . fill in the known values

... ∠CAE = 52° . . . . . . . . subtract 30°

2. ∠CPQ is an external angle to the internal angle CPA of ΔCPA. Thus it is the sum of angles PAC and PCA. Since ΔCPA is isosceles, ∠PAC = ∠PCA and each is half of their sum. That is

... ∠PCA = ∠CPQ/2 = 39°

Likewise, ∠CPQ is an external angle to ΔCQB so is the sum of the equal angles QBC and QCB. Then ...

... ∠QCB = ∠CQP/2 = 31°

The sum of angles in ΔCPQ is 180°, so we have ...

... ∠CPQ + ∠CQP + ∠PCQ = 180°

... 78° + 62° + ∠PCQ = 180° . . . fill in the given values

... ∠PCQ = 40° . . . . . . . . . . . . . .subtract 140°

The angle of interest is the sum of ∠ACP, ∠PCQ, and ∠QCB, so is ...

... ∠ACB = 39° + 40° + 31°

... ∠ACB = 110°