Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 796.

## Sunday, August 19, 2012

### Problem 796: Triangle, Altitude, Measurement of Angles

Labels:
altitude,
angle,
measurement,
triangle

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Al prolongar AD intersecta a BC en ángulo recto por ser el ángulo C=90-x. Así D es el ortocentro del triángulo y debe cumplirse que: x+6+x+x+9=90 y por tanto x=25°.

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http://img13.imageshack.us/img13/6435/problem796.png

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Quadrilateral AHEB is cyclic => AE ⊥ BC

Quadrilateral ACEF is cyclic => CF ⊥ AB

In AHEB angle BAE= ½(180-4x-12)=84-2x …….(1)

In ACEF angle BAE=x+9…… (2)

From (1) and (2) we have 84-2x=x+9 => x=25

Angle HDC = 2x+9

ReplyDeleteBy comparing triangle ADH and DHC, we have HC/AH = tan(2x+9)*tan(x)

By comparing traingel ABH and HBC, we have HC/AH = tan(x) / tan (x+6)

Therefore, tan(2x+9)*tan(x) = tan(x) / tan (x+6) => tan(2x+9)*tan(x+6) = 1

Using product to sum, of sine and consine, we have

cos((2x+9)-(x+6)) - cos((2x+9)+(x+6)) = cos((2x+9)-(x+6)) + cos((2x+9)+(x+6))

cos((2x+9)+(x+6))=0

3x + 15 = 90

x = 25.

Extend AD to meet BC at E. Each of the angles HAE, HBE being x, the 4 points H, A, B, E are con-cyclic. It then follows that angle BDA = BHA = 90. So D is the ortho-center of triangle ABC and CD is perpendicular to AB. Hence the sum of the angles BAE and ABE is 90 degrees, which means x + 9 + x + x + 6 = 90, 3x = 75, x = 25 degrees.

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