Question

w give me solution of this 3 questions question guys please help me

Answer 1

for the third probl: (see diagram).

Draw a circle with AC as the diameter (O is midpoint of AC). The circle will pass through  B and D. Reason is that angle in a semicircle is a right angle.

Now the chord (or arc) CD makes equal angles on the circle, we know this. Hence,  ∠CBD = ∠CAD.

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probl 2:  see diagram.

Circles on two sides intersect on the third.

Let ABC be a triangle (let BC be along horizontal). Draw a circle with AB as  the diameter intersecting AC at D. ∠BDA = 90° (angle in semicircle).

Draw a circle with BC as the diameter. Let it intersect AC at E  and the first circle at F.  Let O be the midpoint of AB and P be the midpoint of BC.
∠BEC = 90° as it is in a semicircle.

We know OP is the line joining midpoints. So it is parallel to AC. 
Line BF is the common chord. We know it is perpendicular to line joining centers. So  BF ⊥ OP,  so   BF  ⊥  AC.

We see that  BD ⊥ AC ,   BE ⊥ AC,  BF ⊥ AC
Since  B is common among all these above,  D, E and F are the same point.
Hence the circles intersect on the third side.

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the first problem   see diagram.

we use the principles of similar triangles.

ΔASC and ΔPSB are similar, as ∠CAB and ∠CPB are equal (angles in a segment made by chord/arc CB). ∠PSB = ∠ASC (vertical angles).
Hence,  ∠PBS = ∠ACP.   -- (1)

∠PBS = ∠DBT (vertical angles).    -- -(2)

∠BDC = ∠BQC  (angles in the same segment made by arc BC).
∠DTB = ∠CTQ.
So ΔDTB & ΔQTC are similar.

So angle ∠QCD = ∠DBT   --- (3)

Hence,  ∠ACP = ∠QCD    by (1), (2) and (3).