Question

Suppose that in triangle ABC, D is a point with B-D-C such that

AD is congruent to CD. Show that the measure of angle ACB is less
than or equal to half the measure of angle ADC.

Answer 1

Using the Isosceles Triangle Theorem and the properties of exterior angles, it is shown that the measure of angle ACB is less than or equal to half the measure of angle ADC in triangle ABC.

Step-by-step explanation:

To show that the measure of angle ACB is less than or equal to half the measure of angle ADC, we start by noting that triangle ADC has two sides AD and CD that are congruent. By the Isosceles Triangle Theorem, angles ADC and CAD are congruent. This implies that angle ADC is twice the measurement of angle CAD.

Looking at triangle ABC, we know that angle ACB is an exterior angle to triangle ADB, so angle ACB must be less than or equal to the sum of the remote interior angles CAD and ADB. Since ADB is part of ADC, and angle CAD is half the measure of angle ADC, it follows that angle ACB is less than or equal to half the measure of angle ADC.

Therefore, the measure of angle ACB is less than or equal to half the measure of angle ADC because ACB is an exterior angle, and ADC is twice the measure of its corresponding interior angle CAD due to the congruence of AD and CD.