Mathematics
In the given figure, AB = AD = DC = PB and ∠DBC = x°. Determine, in terms of x :
(i) ∠ABD, (ii) ∠APB.
Hence or otherwise, prove that AP is parallel to DB.
Answer
(i) Join AC and BD.
∠DAC = ∠DBC = x° [Angles in the same segment are equal]
∠DCA = ∠DAC = x° [As angles opposite to equal sides are equal]
Also, we have
∠ABD = ∠DCA = x° [Angles in the same segment are equal]
Hence, ∠ABD = x°.
(ii) In ∆ABP
⇒ Ext. ∠ABC = ∠BAP + ∠APB
But, ∠BAP = ∠APB [As angles opposite to equal sides are equal]
⇒ 2x° = ∠APB + ∠APB = 2∠APB
⇒ 2∠APB = 2x°
⇒ ∠APB = x°
Hence, ∠APB = x°.
Thus, ∠APB = ∠DBC = x° [These are corresponding angles]
Hence, proved that AP || DB.
Related Questions
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE = 20°; determine ∠BCD. Given reason in support of your answer.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC = 80°. Calculate :
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC.
In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.
In the given figure, AB is the diameter of the circle with centre O.
If ∠ADC = 32°, find angle BOC.
