2

Let $ABCD$ be a square and $T$ a random point inside it. Let $A1,B1,C1,D1$ be points such that they belong to the lines $AT, BT, CT,$ and $DT$ respectively.

Prove that $$\lvert A1B1\rvert\cdot\lvert C1D1\rvert=\lvert A1D1\rvert\cdot\lvert B1C1\rvert$$

A rough sketch of the situation

Note: I can't use trigonometric identities and such. The proof should be as simple as possible, possibly using triangle congruence/similarity, theorems invloving the circle and rotation and the simplest facts involving quadrangles or triangles.

What I've observed is that the identity I'm supposed to prove would imply that $$\frac{\lvert A1B1\rvert}{\lvert A1D1\rvert}=\frac{\lvert B1C1\rvert}{\lvert C1D1\rvert}$$ which would imply congruence of $\triangle A1B1C1$ and $\triangle A1D1C1$ which, honestly, makes no sense to me. Thank you in advance!

mathbbandstuff
  • 657
  • 3
  • 15
  • It shouldn't make sense, because it's not true. Ratios may be equal without the underlying triangles being congruent, or even similar. For instance, just because a triangle with side lengths $3,4,5$ is right, does not mean that any triangle with a ratio of $\frac34$ between two of its sides had a right angle. – Arthur May 04 '17 at 15:45

1 Answers1

1

Let $a$ be the side of the square.

Note that $\triangle TC_1D_1\sim \triangle TDC$.

$$\frac{C_1D_1}{a}=\frac{TC_1}{TD}=\frac{TD_1}{TC}$$

Similarly, we have

$$\frac{D_1A_1}{a}=\frac{TD_1}{TA}=\frac{TA_1}{TD}$$

$$\frac{A_1B_1}{a}=\frac{TA_1}{TB}=\frac{TB_1}{TA}$$

$$\frac{B_1C_1}{a}=\frac{TB_1}{TC}=\frac{TC_1}{TB}$$

Therefore

$$\frac{A_1B_1}{a}\cdot\frac{C_1D_1}{a}=\frac{TA_1}{TB}\cdot\frac{TC_1}{TD}$$

$$\frac{D_1A_1}{a}\cdot\frac{B_1C_1}{a}=\frac{TA_1}{TD}\cdot\frac{TC_1}{TB}$$

The result follows.

CY Aries
  • 23,393