In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If $∠ C B E = 100^{\circ} t h e n ∠ C D F = ?$

Open in App

Solution

In a cyclic quadrilateral ABCD, we have:
Interior opposite angle, $∠ A D C$ = exterior $∠ C B E$ = 100°
∴ $∠ C D F$ = (180° - $∠ A D C$ ) = (180° - 100°) = 80° (Linear pair)

Suggest Corrections

Similar questions

In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If $∠ C B E = 100^{\circ} t h e n ∠ C D F = ?$

(a) $100^{\circ}$

(b) $80^{\circ}$

(d) $90^{\circ}$

Q. In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If ∠CBF = 130° and ∠CDE = x°, find the value of x.

In the figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If $∠ C B F = 130^{\circ} a n d ∠ C D E = x^{\circ},$ find the value of x.
(2 Marks)

Q. In the figure, ABCD is a parallelogram is produced to E such that

A B = B E

. AD produced to F such that

A D = D F

. Show that

△ F C D ≅ △ C B E

Q. In the figure given above the sides, AB and AC of

△

ABC are produced to points E and D respectively. If bisectors BO and CO of

∠ C B E

and

∠ B C D

respectively meet at point O, then prove that

∠ B O C = 90^{o} - \frac{1}{2} ∠ B A C

Join BYJU'S Learning Program

In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If ∠CBE=100∘ then ∠CDF=?

In a cyclic quadrilateral ABCD, we have: Interior opposite angle, ∠ADC= exterior ∠CBE = 100° ∴ ∠CDF = (180° - ∠ADC) = (180° - 100°) = 80° (Linear pair)

In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If $∠ C B E = 100^{\circ} t h e n ∠ C D F = ?$

In a cyclic quadrilateral ABCD, we have:
Interior opposite angle, $∠ A D C$ = exterior $∠ C B E$ = 100°
∴ $∠ C D F$ = (180° - $∠ A D C$ ) = (180° - 100°) = 80° (Linear pair)