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.