PLEASEEEEE HEEEEEELPPPPPPPPPPPPPPPSuppose ABCD is a rhombus such that the angle bisector of ∠ABD meets AD at point K. Prove that m∠AKB = 3m∠ABK.
Accepted Solution
A:
1. Let m∠ABK=x°. Since line BK bisects ∠ABD, thenm∠ABK=m∠KBD=x°.Also m∠ABD=m∠ABK+m∠KBD=2x°.2. The diagonal BD of rhombus ABCD bisects ∠ABC, then m∠ABD=m∠DBC=2x°.This gives you thatm∠ABC=4x°.3. Angles A and B are supplementary, som∠A+m∠B=180°,m∠A=180°-4x°.4. Consider triangle ABK. The sum of the measures of interior angles in triangle is always 180°, thusm∠A+m∠ABK+m∠AKB=180°,m∠AKB=180°-x°-(180°-4x°),m∠AKB=3x°=3m∠ABK.