The Problems of Ukrainian Authors

Question

Consider the acute ABC and point D on the side AB. Let's denote the center of the circumscribed circle around the ACD as P and the center of the circumscribed circle around the BDC as Q. Prove that triangles ABC and DPQ are similar. (Bogdan Rublyov) FIGURE_0

Accepted Answer

Let's denote the radius of the circumscribed circle around the

△ A C D

as

R_{1}

, the radius of the circumscribed circle around the

△ B D C

as

R_{2}

. So, if

C D = l

then by the law of sines (Fig.47):

\frac{l}{sin β} = 2 R_{1} \Rightarrow sin β = \frac{l}{2 R _{1}} .

Herewith,

cos φ = \frac{D E}{D Q} = \frac{l}{2 R _{1}} = sin β

. Therefore, from the acute triangle

△ A B C

φ = \frac{π}{2} - β

, and similarly

ψ = \frac{π}{2} - α

. Therefore

∠ P D Q = φ + ψ = \frac{π}{2} - α + \frac{π}{2} - β = π - α - β = γ = ∠ A C B

.

Fig. 47

Moreover, from the following equation:

\frac{Q D}{P D} = \frac{R _{1}}{R _{2}} = \frac{l}{2 sin β} \cdot \frac{2 sin α}{l} = \frac{sin α}{sin β} = \frac{a}{b} = \frac{C B}{A C}

the sides about the equal angles of our triangles are proportional, so triangles

A B C

and

D P Q

are similar for any point

D \in A B

. Points

A, B

do not satisfy the condition, otherwise one of the

△ A C D

or

△ B D C

degenerates to a line segment.