# In triangle ABC, AB=52, BC=56, CA=60. Let D be the foot of the altitude from A, and let E be the intersection of the internal bisector of the internal angle bisector of angle BAC with BC. Find DE.

DE=6. We are given `Delta ABC` with AB=52, BC=56, and AC=60. The intersection of the altitude from A intersects BC at D, and the interior angle bisector of ` /_ BAC` intersects BC at E. We are asked to find DE. (See attachment for diagram.)

Since AD is an altitude, it forms right angles between AD and BC. Let BD=x and then CD=56-x. Using the Pythagorean theorem on `Delta ADB` we get `x^2+(AD)^2=52^2` . Applying the Pythagorean theorem to `Delta ADC` we get `(56-x)^2+(AD)^2=60^2` . Solving both equations for `(AD)^2` we set the expressions equal to each other to get the following:

`(56-x)^2-x^2=60^2-52^2`
`3136-112x+x^2-x^2=896`
`112x=2240`
`x=BD=20` and CD=36.

Since AE is an angle bisector of A, it divides BC proportionally to the other legs so that we get the proportion `(AB)/(BE)=(AC)/(EC)` or `52/(BE)=60/(EC)` .

Let BE=y so that EC=56-y.

`52/y=60/(56-y)` ; use the means-extremes property (cross-multiply) to get
`60y=52(56-y)`
`60y=2912-52y`
`112y=2912`
`y=BE=26` and CE=30.

BD+DE+CE=56 or 20+DE=30=56 ==> DE=6.

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