Question

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AB and CE bisects ZBCD. Prove that:
(i) AE = AD
(ii) DE bisects ZADC
(iii) ZDEC = 90°.
Hint: ZBEC = ZECD = ZECB = EB = BC = AE = AD.
... ZADE = ZAED = ZEDC.
ZADC + ZBCD = 180°
(ZADC) + 1 / 2 (ZBCD) = 90° = ZEDC + ZDCE = 90°]
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Answer 1

Answer:

Answer is mentioned in the explanation down

STEP BY STEP EXPLANATION

Given CE bisects angle BCD

that means angle DCE=angle CBE

Now AB is parallel to CD AND CE is transversal

so, angle DCE =angle CEB (alternate interior angles)

given angle ADE=angle AED=angle EDC

that means angle ADE=angle EDC that means

DE bisects angle ADC

If we consider DEC as a triangle 90°+angle DEC =180°

that means angle DEC =90°

in hint they gave AE=AD

HENCE PROVED