# Difference between revisions of "2020 USOJMO Problems/Problem 4"

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+ | Let <math>G</math> be on <math>AB</math> such that <math>GD \perp DB</math>, and <math>H = GD \cap EB</math>. Then <math>\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}</math> is the orthic triangle of <math>\triangle{HGB}</math>. Thus, <math>F</math> is the midpoint of <math>GB</math> and lies on the <math>\perp</math> bisector of <math>DB</math>. |

## Latest revision as of 13:27, 19 April 2021

## Problem

Let be a convex quadrilateral inscribed in a circle and satisfying . Points and are chosen on sides and such that and . Prove that .

## Solution

Let be the intersection of and and be the intersection of and .

**Claim: **

By Pascal's on , we see that the intersection of and , , and are collinear. Since , we know that as well.

Note that since all cyclic trapezoids are isosceles, . Since and , we know that , from which we have that is an isosceles trapezoid and . It follows that , so is an isosceles trapezoid, from which , as desired.

## Solution 2

Let , and let . Now let and .

From and , we have so . From cyclic quadrilateral ABCD, . Since , .

Now from cyclic quadrilateral ABC and we have . Thus F, A, D, and E are concyclic, and Let this be statement 1.

Now since , triangle ABC gives us . Thus , or .

Right triangle BHC gives , and implies

Now triangle BGE gives . But , so . Using triangle FGD and statement 1 gives

Thus, , so as desired.

~MortemEtInteritum

## Solution 3 (Angle-Chasing)

Proving that is equivalent to proving that . Note that because quadrilateral is cyclic. Also note that because . , which follows from the facts that and , implies that . Thus, we would like to prove that triangle is similar to triangle . In order for this to be true, then must equal which implies that must equal . In order for this to be true, then quadrilateral must be cyclic. Using the fact that , we get that , and that , and thus we have proved that quadrilateral is cyclic. Therefore, triangle is similar to isosceles triangle from AA and thus .

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## Solution 4

BE is perpendicular bisector of AC, so . FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. . Hence, , .

Mathdummy

## Solution 5

Let be on such that , and . Then is the orthic triangle of . Thus, is the midpoint of and lies on the bisector of .