⇒CD=h‌cot‌3‌α . . . (i) In △EBD, ‌tan‌2‌α=‌
h
BD
⇒ ‌BD=h‌cot‌2‌α . . . (ii) In △EAD tan‌α=‌
h
AD
⇒ AD=h‌cot‌α . . . (iii)
From Eqs. (ii) and (iii), AD−BD‌=h‌cot‌α−h‌cot‌2‌α AB‌=h(cot‌α−cot‌2‌α) . . . (iv) From Eqs. (i) and (ii), BD−CD=h‌cot‌2‌α−h‌cot‌3‌α BC=h(cot‌2‌α−cot‌3‌α) . . . (v) From Eqs. (iv) and (v), ‌‌