## Tuesday, August 6, 2024

### Geometry Problem 1576: Congruency of Segments in Triangle ABC with Angles 30 and 20 Degrees and an Interior Cevian

Challenging Geometry Puzzle: Problem 1576. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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1. Trigonometry Solution

Let BD = CD = p and let AD = d

From Triangle ABD,
p/sin30 = d/sin70 ................(1)

From Triangle BCD,
a = 2p cos20 = 2p sin70 ............(2)

(1) X (2) gives,
ap/sin30 = 2pd
So a = 2dsin30 = d

Sumith Peiris
Moratuwa
Sri Lanka

2. Pure Geometry Solution

Extend BD to E such that BD = DE, let BD = DC = DE = p
Let DF be perpendicular to AB, F on AB extended

Note that ADF is a 30-60-90 Triangle and so DF = d/2

Now < BCE = 90 and Triangles BFD and BCE are similar

Hence DF / BD = BC / BE which gives
(d/2)/p = a/ (2p)

Therefore d = a, that is AD = BC

Sumith Peiris
Moratuwa
Sri Lanka