Find angle in a semicircle.

Additional information about inscribed angles and inscribed/cyclic quadrilaterals could be found here:

[اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .

[panPetrff]

I tried it on my own. I noticed that angle B = 90-26=64° equals 180-2*58 = 180 - 116
So we have (after completing circle, extending B→E→F and joining DF) a cyclic quadrilateral DCBF where angle <ADC = <EDF = 58° and as DE lies on the diameter, triangles EDC and EDF are congruent. Then angle <EFD = <BFD equals angle <ECD = x
Opposite vertices C and F have supplementary angles 180° = (x+26) + x , finaly 2x=180-26=154 ...so x=154/2=77°

[jimlocke]

You can get to x = 77° in fewer steps by proving that ΔCGD and ΔFGD are congruent and proceeding from there. At about 3:10, <FCD = <FBD = 32° is proved. By the same reasoning, but using a different common chord, <CBD = <CFD = 32°. Then, in the video, <CGD is found to equal 90°. AD and CF are intersecting line segments, so all 4 angles are 90°. Therefore, <DGF = 90°, and by sum of interior angles of a triangle = 180°, <GDF = 58°. Now, ΔCGD and ΔFGD are congruent by angle-side-angle (90° - common side GD - 58°). ΔEGF and ΔEGC are congruent by side angle side (common side EG - 90° - equal sides FG and CG). In ΔCEF, <CEF is a right angle and sides CE and CF are equal in length (by ΔEGF and ΔEGC being congruent). Thus, ΔCEF is an isosceles right triangle, and <ECF = EFC = 45°. <ECF = ECG because it is the same angle. x = <ECG + <GCD = 45° + 32° = 77°

[kevinmorgan]

You can also show that the semi circle with BC as a diameter must pass through the centre of the big semi circle with AD as diameter. Then some jiggling with the numbers yields the same result.

[HassanLakiss]

Thank you for a nice challenging question and your explanation was easy to follow.With respect, BEF is a straight line and hence angle CEF is 90, since BEC is 90. Hence, each purple angle is 45. Incidentally, your presentation is very nice and colourful. What software are you using to draw and write with? Thank you.

[Mathematician]

Ah this was nice. I solved it with ease, because I had done a similar problem before. Then it took some time. I was not good at geometry then.

The problem was, ADKB is a semicircle, (A, D, K, B are in clockwise order) P is a point On diameter AB such that angle KPB=angle APD, prove that ADP is similar to BPK.

[ronaldovieirafragosoronald]

Much more simple: quadrilateral BCDF: < x+26 and < BFD X x+26 +x =180 x= 77º