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From a point $P$ outside a circle, draw two tangents to the circle touching at points $A$ and $B$. Draw a sectant line intersecting the circle at points $C$ and $D$, with $C$ between $P$ and $D$. Choose point $Q$ on the chord $CD$ such that $\angle DAQ=\angle PBC$. Prove that $\angle DBQ=\angle PAC$.

My Attempt enter image description here

I have somehow made a figure but could not solve it. Please help me with this..

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1 Answer 1

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Let O be the center of the circle ACBD.

enter image description here

Fact-1: All the same color coded angles are equal.

Fact-2: The red dotted circle will pass through A,P, B, O.

Fact-3: $\angle purple1 = \angle green + \angle red$, a standard result from tangent properties.

By considering $\triangle ADQ$, $\angle blue = \angle green + \angle red$.

Since $\angle purple1 = \angle blue$, we say that A, P, B, Q, O are con-cyclic.

Then, $\angle black = \angle purple2 = \angle purple1$

That means $\angle green + \angle yellow = \angle green + \angle red$.

Result follows.

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  • $\begingroup$ why are those angles (same colored) equal? $\endgroup$
    – pi-π
    Feb 4, 2017 at 16:18
  • $\begingroup$ @NeWtoN The two reds are equal because of "angle in alternate segment". The equality of the greens can be deduced the same way. the purples are equal because of "tangent properties". $\endgroup$
    – Mick
    Feb 4, 2017 at 16:27
  • $\begingroup$ @@Thanks mick,, could you please help me with this question too.. math.stackexchange.com/questions/2128331/… $\endgroup$
    – pi-π
    Feb 4, 2017 at 16:37
  • $\begingroup$ @NeWtoN I'll try but I haven't seen this type of question before. $\endgroup$
    – Mick
    Feb 4, 2017 at 19:32
  • $\begingroup$ Ok..for sure. You may try $\endgroup$
    – pi-π
    Feb 5, 2017 at 1:33

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