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.