Mathematics
In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.
Related Questions
In △ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that △CAE is isosceles.
In the adjoining figure, ABC is a right angled triangle at B. ADEC and BCFG are squares. Prove that AF = BE.
In the adjoining figure, AB = AC, D is a point in the interior of △ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of △ABC.
In the adjoining figure, PQ || BA and RS || CA. If BP = RC, prove that:
(i) △BSR ≅ △PQC
(ii) BS = PQ
(iii) RS = CQ.
