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Topic: "Any point P on the hypotenuse BC of isosceles right triangle ABC intersects with point P to make PD perpendicular to AB and PE perpendicular to AC?" , link CD, pay PE at point M; Connect BE to PD at point n.

Verification: PM=PN "

Prove:

Because PE⊥AC, PD⊥AB, angle cab right angle,

Therefore, the quadrilateral PEAD is a rectangle.

Available: PD=AE? ; EP = AD

Because △CAB is an isosceles right triangle and AB=AC.

Therefore, ce = EP = adPD=AE=BD.

Rt△PMC∽Rt△EMC

Available: PM/EM=PD/CE

PM? /? (PE-PM)? =? PD? /? (AC acoustic emission)

PM? /? (PE-PM)? =? PD? /? (AC PD)

(PE-PM)? /? PM? =? (AC-PD)? /? Parkinson's disease

PE? /? PM? =? AC? /? Parkinson's disease

PM? =? PE*PD? /? Alternating current

Similarly: Rt△PEN∽Rt△DBN

Available: PN? /? DN? =? PE? /? decibel

PN? /? (PD-PN)? =? PE? /? (AB-AD)

PN? /? (PD-PN)? =? PE? /? (AB-PE)

(PD-PN)? /? PN? =? (AB-PE)? /? PE

PD? /? PN? =? AB? /? PE

And said: PN? =? PD*PE? /? ab blood type

Take it to the front: PM? =? PE*PD? /? Alternating current

Because, AB=AC

So, PM=PN.

S.H.I.E.L.D.